How to find parameters from an equation the question could be "stupid" but i don't know if it is feasible or not, please don't kill me :)
EDIT WITH NEW FORMULAS!
I have an equation like this: (unfortunatly in my first Q&A i cannot upload images for "spam" reason, I post latex version of the formula hoping it is understandable. Otherwise which kind of representation can i use?)
$$N_c = \sum_{i=1}^{max} \bigg( \frac{1}{i^s H} * N_u * i\bigg)$$
where, $H = \sum_{i=1}^{max} \frac{1}{i^s}$
If all the variables are known except one: 


*

*"$max$" is UNKNOWN


it is possible to find the "$max$" parameter? Or it is mathematically impossible?
Thank you very much.
ADDED (based on Ross Millikan answer):
Practical example based on you approximations to verify the correcteness of "H" formula. If i set:


*

*$s = 0.5$

*$max = 500$ (in this case i know also the "max" value, i want only to verify the integral)


we have:
$$H=\sum_{i=j}^{m} \frac{1}{j^s}\approx \int_1^{m} x^{-s} \; dx=\frac{1}{(1-s)x^{1-s}}\mid _1^{m}=\frac{1}{1-s}\left(1-\frac{1}{m^{1-s}}\right) = $$ 
$$ = \frac{1}{1-0.5}\left(1-\frac{1}{500^{1-0.5}}\right) = 2 \times 0.95 = 1.91$$
But if I calculate the original summatory we obtain: 
$$H=\sum_{i=j}^{m} \frac{1}{j^s} = 43.28$$
I think there is something strange in the integral's "evolution". I think we can expand it in this way:
$$H=\sum_{i=j}^{m} \frac{1}{j^s}\approx \int_1^{m} x^{-s} \; dx=\frac{x^{1-s}}{(1-s)}\mid _1^{m}= 
\frac{500^{0.5}}{1-0.5} - \frac{1^{0.5}}{1-0.5} = 44.72 - 2 = 42.72$$
that is a better approximation. What do you think? I don't know if it is correct, this is a "new world" for me!
 A: Unless there is unstated dependence on $i$, you can distribute out $H$ and $N_u$.  If so you have $$N_c = \frac{N_u}{H}\sum_{i=1}^{max} \frac{1}{i^s}$$  This is a series that can be summed in terms of generalized harmonic numbers, but I am not sure that helps a lot.  You can also approximate the sum by an integral $$\sum_{i=1}^{max} \frac{1}{i^s}\approx \int_1^{max} x^{-s} \; dx=\frac{1}{(1-s)}x^{1-s}\bigg| _1^{max}=\frac{1}{(1-s)}(max^{1-s}-1)$$and be pretty close.  If you want the exact value you could then search.
Added:  for the new version, your $H$ is just the same as the sum above, but then you need to sum again.  I would presume the sums are over two different variables, so I will change the $H$ sum to go over $j$ and let $m$=max: $$H=\sum_{i=j}^{m} \frac{1}{j^s}\approx \int_1^{m} x^{-s} \; dx=\frac{x^{1-s}}{1-s}\bigg| _1^{m}=\frac{1}{1-s}\left(m^{1-s}-1\right)$$
Now your new $$N_c=\frac{N_u}{H}\sum_{i=1}^m\frac{i}{i^s}=\frac{N_u}{H}\sum_{i=1}^m\frac{1}{i^{s-1}}$$  and we can use the same approximation, as the sum is the same except for being $s-1$:  $$N_c\approx N_u\frac{1-s}{2-s}\frac{m^{2-s}-1}{m^{1-s}-1}$$
